Edexcel FP1 — Question 6 6 marks

Exam BoardEdexcel
ModuleFP1 (Further Pure Mathematics 1)
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSequences and series, recurrence and convergence
TypeSum from n+1 to 2n or similar range
DifficultyStandard +0.3 This is a straightforward application of standard summation formulas. Part (a) requires algebraic manipulation of known results for Σr and Σr³, which is routine for FP1 students. Part (b) uses the 'hence' technique of subtracting two evaluations, a standard exam pattern. While it involves multiple steps, no novel insight is required—just careful execution of familiar techniques.
Spec4.06a Summation formulae: sum of r, r^2, r^3

6. (a) Use the standard results for \(\sum _ { r = 1 } ^ { n } r\) and for \(\sum _ { r = 1 } ^ { n } r ^ { 3 }\) to show that, for all positive integers \(n\), $$\sum _ { r = 1 } ^ { n } r \left( 2 r ^ { 2 } - 6 \right) = \frac { 1 } { 2 } n ( n + 1 ) ( n + 3 ) ( n - 2 ) .$$ (b) Hence calculate the value of \(\sum _ { r = 10 } ^ { 50 } r \left( 2 r ^ { 2 } - 6 \right)\).

Question 6:
Part (a)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\sum 2r^3 - 6r\)M1
\(2\dfrac{n^2}{4}(n+1)^2 - 6\dfrac{n}{2}(n+1)\)A1
\(= \dfrac{n}{2}(n+1)\left[n(n+1)-6\right]\)M1
\(= \dfrac{n}{2}(n+1)\left[n^2+n-6\right]\)
\(= \dfrac{n}{2}(n+1)(n+3)(n-2)\)A1 cos (4 marks) marked with (*)
Part (b)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(f(50) - f(9) = 3243600 - 3780\)M1
\(= 3239820\)A1 (2 marks)
(6 marks total)
# Question 6:

## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sum 2r^3 - 6r$ | M1 | |
| $2\dfrac{n^2}{4}(n+1)^2 - 6\dfrac{n}{2}(n+1)$ | A1 | |
| $= \dfrac{n}{2}(n+1)\left[n(n+1)-6\right]$ | M1 | |
| $= \dfrac{n}{2}(n+1)\left[n^2+n-6\right]$ | | |
| $= \dfrac{n}{2}(n+1)(n+3)(n-2)$ | A1 cos | (4 marks) marked with (*) |

## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(50) - f(9) = 3243600 - 3780$ | M1 | |
| $= 3239820$ | A1 | (2 marks) |
| **(6 marks total)** | | |

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6. (a) Use the standard results for $\sum _ { r = 1 } ^ { n } r$ and for $\sum _ { r = 1 } ^ { n } r ^ { 3 }$ to show that, for all positive integers $n$,

$$\sum _ { r = 1 } ^ { n } r \left( 2 r ^ { 2 } - 6 \right) = \frac { 1 } { 2 } n ( n + 1 ) ( n + 3 ) ( n - 2 ) .$$

(b) Hence calculate the value of $\sum _ { r = 10 } ^ { 50 } r \left( 2 r ^ { 2 } - 6 \right)$.\\

\hfill \mbox{\textit{Edexcel FP1  Q6 [6]}}